For which $0^\circ < \theta < 500^\circ$ is $3\cos \theta + 4\sin \theta > 2$?

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\(3\cos \theta + 4\sin \theta\) may be written in the form \(r\cos(\theta - \alpha)\), where \(r > 0\).

1. Calculate the value of \(r\), and show that one value of \(\alpha\) is approximately \(53.1^\circ\).

Can we expand the expression \(\cos(\theta - \alpha)\)?

2. Hence show that one solution of \(3\cos \theta + 4\sin \theta = 2\) is approximately \(120^\circ\).

Can we use the result from part (a)?

3. State all the other solutions for \(0^\circ \leq \theta \leq 500^\circ.\)

How many solutions do you expect in the interval \(0\leq\theta<360^\circ\)?

How can we locate further solutions?

4. Hence give the positive values of \(\theta\) less than \(500^\circ\) for which \(3\cos \theta + 4\sin \theta > 2\).

A sketch graph of \(y = 3\cos \theta + 4\sin \theta\) might be helpful. What happens at each of the solutions found in part (c)?